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# Vapour pressure question

## Vapour pressure question

(OP)
Hi all

I feel like I am losing my mind. There's something that's confusing me:

We are filling LPG tankers from LPG spheres. The spheres are at about 6 bar, 50/50 mixture of propane/butane. The LPG exits the bottom of the sphere and gets pumped by vane (positive displacement) pumps through the metering skid and into the tanker. Keep in mind, there is no vapour return line. While loading, I would expect the pressure of the tanker to not exceed the vapour pressure until the tanker is liquid full. This is not the case - it reaches 14 bar. At 14 bar, the LPG should be subcooled and the vapour pressure should be even lower. Is it possible that we are just pumping faster than the vapour can condense? (We are pumping at 24t/h, but as we reach near 14bar, it slows down to 1t/h - I would then have expected the pressure in the tanker to drop if this was the case).

My question is why does the tanker reach 14 bar if the vapour pressure of a 50/50 mixture at 20C is about 4 bar and the tanker is only 80% full?

Any help would be appreciated :)

### RE: Vapour pressure question

It's getting warmer?

### RE: Vapour pressure question

(OP)
No, it doesn't get warmer, in fact due to Joule-Thomson expansion, it gets colder.

### RE: Vapour pressure question

IMHO, one possible reason could be an "inert gas" present in the tanker becoming compressed by the liquid LPG.

### RE: Vapour pressure question

(OP)
AH! Thanks 25362. That is a very plausible answer! But as far as VLE is concerned (excl inerts), there's no ways this could be possible, right?

### RE: Vapour pressure question

What pressure and temperature does the tanker start at?

What temp is your LPG at?

Does the LPG initially flash off when delivery starts?

How have you dealt with the compression of the gas in the tanker in your thought process if there is no return vapour path?

a sketch with all your initial start pressures, pumps pressure controllers etc is always useful.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

### RE: Vapour pressure question

(OP)
Tanker starts at about 25C and ranges from 2bar - 10bar (most recent was 3.5bar).

The LPG is stored in the sphere at 25C, 5.9bar.

Unfortunately I don't know if it flashes, they have a rotameter to check the level. I do think that the first bit flashes though.

I only commissioned the plant, I did not design it. I would always put a vapour return line, but in this design they did not have one.

I'm not sure how necessary the diagram is - I can make you a sketch if you need one though (can you PM on this forum?)

If I can try and make this as basic as possible, I wanted to know if there is some way (besides inerts) that the pressure in a tanker can be higher than the vapour pressure? As far as I know, the pressure inside the tanker should be equivalent to the vapour pressure (if assuming no inerts). Is there something that I am missing?

Thanks for the interest LittleInch, let me know if you still want that sketch :)

### RE: Vapour pressure question

Can you tell us whether the "abnormal" high tanker pressure is permanent or temporary and coming slowly down?

### RE: Vapour pressure question

(OP)
I can't - I was wondering about that today as well, maybe it hasn't reached equilibrium. We have 4 tankers coming back tomorrow and I should recognise a couple of the drivers so I'll ask them :)

### RE: Vapour pressure question

If the pressure comes down it is possible that lighter-than-LPG hydrocarbons either flashed upon filling the tankers, or otherwise remnant from previous operations, gradually redissolve.

### RE: Vapour pressure question

(OP)
Would the lighter-than-LPG hydrocarbons be able to exert that much pressure? The way I see it, the vapour pressure exerted by a single component in a mixture is approx. the mol fraction * Pvap (Raoult's law) and even if there were lighter components, the molar frac should be small enough that it has negligible impact (tell me if you disagree - this is just my opinion).

### RE: Vapour pressure question

I'm sure others can explain this better than me, but I'm not surprised at all. You need to remember that the vapour pressure line does not mean that the instant LPG reaches above that pressure it instantly returns to a liquid. There is a lot of heat transfer / enthalphy to take place to allow this which you're not taking into consideration.

From the basic info supplied, it looks to me like your LPG is initially flashing into the tanker, fairly rapidly pushing it's pressure to about 5 - 6barg. Then the rest of the liquid arrives fairly quickly into your closed vessel (tanker) and starts to compress this vapour. If you start at 6 bara and about 10% full and then go to 80% full, you get about 15bara. The vapour is probably a misty gas, but without some cooling to get it to change phase it won't magically transform itself to liquid again. If you have a vapour return line set at 6 barg then you wouldn't have this issue.

Your comment that at 14 barg the liquid would be subcooled doesn't look to be correct - where has the heat gone to make it supercooled? If you lowered the gas pressure in the tanker to 6 barg, you probably wouldn't get any vapourisation but that doesn't mean it's supercooled.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

### RE: Vapour pressure question

(OP)
Hi LittleInch

Thanks for your patience. I think I understand what you're saying - basically the gas is in transient state and in a sort of aerosol? How did you calc the 15bara, IGL? I think the answer from the drivers tomorrow will clear things up.

I wish the guys had installed a vapour return line, when we reach about 80% of capacity, the flow rate drops from 24t/h to about 0.8t/h. This is why I started this thread :)

To answer your question - the heat has been lost to Joule-Thomson expansion. But this is only initially. The thing that confuses me is that at 25C, the Pvap of propane is 8.8bar and of butane is about 1.6bar. That's nowhere near the 14bar we see. When I say subcooled, all I mean is that it is at a higher pressure than saturation pressure at the temp (or a lower temp than the saturation temp at that pressure). Does this make sense?

### RE: Vapour pressure question

I didn't mean the low mol fraction of light HC's exert a high partial pressure (Raoult's law). I intended to explain, unfortunately w/o success, that whatever the composition of the gas present in the tankers prior to the filling operation (probably an enriched propane mixture) becomes compressed by the entering liquid together with the flashed vapors.

Redissolution (attainment of equilibrium) takes time. That was the reason of my question about a possible reduction of the "abnormal" pressure after a lapse of time.

### RE: Vapour pressure question

Yes a fairly basic gas law calculation. Your high pressure, IMO, is simply because you are compressing the gas in the tanker, not because the LPG is boiling and releasing it. This pressure can be anything greater than the vapour pressure. If it's higher than the VP then the amount of liquid LPG turning into gas reduces dramatically. However the opposite (HP gas turning back into liquid) is not necessarily the case unless you cool the gas to make it turn back into liquid.

If you could vent the tankers to maintain them at say 8 barg, you wouldn't have that problem me thinks.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

### RE: Vapour pressure question

Agree with what LittleInch has said. The process appears to be as follows:

• The vapor, a mixture of pre-existing and flashed vapor, is being compressed and somewhat heated by the entering liquid.
• Part of the vapor condenses in contact with the colder liquid surface.
• The released heat of condensation cannot be quickly distributed throughout the entire liquid mass by convection. Therefore the surface temperature is warmer then the bulk below the surface.
• The compression of the vapor and the non-uniform temperature distribution in the liquid causes the pressure in the tanker to exceed the vapor pressure corresponding to the mean liquid temperature.
• After some time equilibrium would be reached, and the pressure would come down to a vapor pressure level.

### RE: Vapour pressure question

(OP)
25362 - Makes sense, I thought that when pressurised, the propane would condense pretty quickly (though I have no grounds to think this happens quickly - I guess this is just something I imagined and never questioned it)

LittleInch - thanks :) Since there's no weigh bridge, we only measure the volumetric flow, if we vented, the guys could argue that we are charging them for gas that they haven't received

25362 - That's a great summary :) I agree with the mechanism

25362 and LittleInch, thanks so much for taking the time to help! I really appreciate it

### RE: Vapour pressure question

1) Small amounts of CO2, N2, O2, CO, C2's may be accumulating in the tank, although probably very little and more of a long term problem.
2) As the gas in the tank is compressed to smaller volume it will increase in temperature and pressure simultaneously. The vent line would ideally be sized to relieve the volume of liquid in gas as the tank fills up to maintain constant pressure and temperature.

### RE: Vapour pressure question

If you instal a vapor return line you will have no measurement.

The tankers should be vapor recovered at the unload point sending you tans with under 2 brag back to fill.

You could load faster if you went to a pump that could put up 3+ brag differential pressure! and have no vapor return line.

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